オリジナル問題集No.1

次の様に数を並べてみると良い。
各行について等比級数の公式を適用する事で直ちに本問題の等式が証明される。

$$\begin{array}{rcl}\sum _{m,n=0}^N{x}^{|m-n|}& =& N+2\sum _{k=1}^{N-1}(N-k)x^k\\ & =& N+2\sum _{k=1}^{N-1}k{x}^{N-k}\\ & =& N+2x^N\sum _{k=1}^{N-1}k{x}^{-k}\\ & =& N+2x^N\frac{1}{1-x^{-1}}(\sum _{k=1}^{N-1}{x}^{-k}-(N-1)x^{-N})\\ & =& N+2x^N\frac{x^{-1}-x^{-N}}{(1-x^{-1}{)}^2}-2(N-1)\frac{1}{1-x^{-1}}\\ & =& N+2\frac{x^{N+1}-x^2}{(x-1{)}^2}-2(N-1)\frac{x}{x-1}\end{array}$$

下の様に積分領域を分割する。
すると与えられた積分は下記の様に書ける。
$${\int }_0^{a}d{x}_1{\int }_0^{b}d{x}_{2}f(x_1)f(x_2)dx={\int }_0^{a}d{x}_1{\int }_0^{\frac{b}{a}{x}_1}d{x}_{2}f(x_1)f(x_2)+{\int }_0^{b}d{x}_2{\int }_0^{\frac{a}{b}{x}_2}d{x}_{1}f(x_1)f(x_2)$$

積分領域

[1]まずは前提になる式を書きます。
$${\int }_0^1\frac{(\log x{)}^m}{1-x}dx=(-1{)}^{m}m!\zeta (m+1)$$
この式の証明は簡単なので省略いたします。
[2]次に$m=2$として二乗を計算します。
$$\begin{array}{rcl}\{{\int }_0^1\frac{(\log x{)}^2}{1-x}dx{\}}^2& =& 4\zeta (3{)}^2\\ & =& 2{\int }_0^1\frac{(\log x{)}^2}{1-x}dx{\int }_0^x\frac{(\log y{)}^2}{1-y}dy(z=\frac{y}{x})\\ & =& 2{\int }_0^1\frac{x(\log x{)}^2}{1-x}dx{\int }_0^1\frac{(\log xz{)}^2}{1-xz}dz\\ & =& 2{\int }_0^1\frac{x(\log x{)}^4}{1-x}dx{\int }_0^1\frac{1}{1-xz}dz+4{\int }_0^1\frac{x(\log x{)}^3}{1-x}dx{\int }_0^1\frac{\log z}{1-xz}dz+2{\int }_0^1\frac{x(\log x{)}^2}{1-x}dx{\int }_0^1\frac{(\log z{)}^2}{1-xz}dz\\ & =& -2{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx+4\sum _{n=0}^{\mathrm{\infty }}{\int }_0^1\frac{x^{n+1}(\log x{)}^3}{1-x}dx{\int }_0^1{z}^n\log zdz+2\sum _{n=0}^{\mathrm{\infty }}{\int }_0^1\frac{x^{n+1}(\log x{)}^2}{1-x}dx{\int }_0^1{z}^n(\log z{)}^{2}dz\\ & =& -2{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx-4\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+1{)}^2}{\int }_0^1\frac{x^{n+1}(\log x{)}^3}{1-x}dx+4\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+1{)}^3}{\int }_0^1\frac{x^{n+1}(\log x{)}^2}{1-x}dx\\ & =& -2{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx+24\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(m+n+2{)}^4(n+1{)}^2}+8\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(m+n+2{)}^3(n+1{)}^3}\\ & =& -2{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx+24\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(m+n{)}^4{n}^2}+8\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(m+n{)}^3{n}^3}\\ & =& -2{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx+24\zeta (2,4)+8\zeta (3,3)\end{array}$$
[3]ゆえに、以下の式を得る。
$$\begin{array}{rcl}{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx& =& 12\zeta (2,4)+4\zeta (3,3)+2\zeta (3{)}^2(\zeta (3,3)=\frac{1}{2}\{\zeta (3{)}^2-\zeta (6)\})\\ & =& 12\zeta (2,4)+4\zeta (3{)}^2-2\zeta (6)\end{array}$$
[4]実はまめひげさんの記事の定理7を参考にすると以下の式を得る。
$$\begin{array}{rcl}{\int }_0^1\frac{(\log x{)}^4\log (1-x)}{1-x}dx& =& -60\zeta (6)+12\{\zeta (2)\zeta (4)+\zeta (3)\zeta (3)+\zeta (4)\zeta (2)\}\\ & =& 24\zeta (2)\zeta (4)+12\zeta (3{)}^2-60\zeta (6)\end{array}$$
ゆえに、以下の式を証明すればいい。
[5]$24\zeta (2)\zeta (4)-60\zeta (6)=-18\zeta (6)$
$$\begin{array}{r}\{\begin{array}{l}\zeta (2)=\frac{{\pi }^2}{6}\\ \zeta (4)=\frac{{\pi }^4}{90}\\ \zeta (6)=\frac{{\pi }^6}{945}\\ \zeta (2)\zeta (4)=\frac{7}{4}\zeta (6)\end{array}\end{array}$$
を用いると
$$\begin{array}{rcl}24\zeta (2)\zeta (4)-60\zeta (6)& =& 42\zeta (6)-60\zeta (6)\\ & =& -18\zeta (6)\end{array}$$
[6]以上をまとめると
$$\begin{array}{r}\{\begin{array}{l}\zeta (2,4)=\frac{2}{3}\zeta (3{)}^2-\frac{4}{3}\zeta (6)\\ \zeta (4,2)=\zeta (2)\zeta (4)-\frac{2}{3}\zeta (3{)}^2+\frac{1}{3}\zeta (6)\end{array}\end{array}$$

$$\begin{array}{rcl}({\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx{)}^2& =& 2{\int }_0^{1}dx{\int }_0^{x}dy(xy{)}^{m-1}\{(1-x)(1-y){\}}^{n-1}(\frac{y}{x}=z)\\ & =& 2{\int }_0^{1}dx{x}^{2m-1}(1-x{)}^{n-1}{\int }_0^{1}dz{z}^{m-1}(1-zx{)}^{n-1}\\ & =& 2\sum _{k=0}^{n-1}(-1{)}^k\left(\begin{array}{c}n-1\\ k\end{array}\right){\int }_0^{1}dx{x}^{k+2m-1}(1-x{)}^{n-1}{\int }_0^{1}dz{z}^{k+m-1}\\ & =& 2\sum _{k=0}^{n-1}(-1{)}^k\frac{\left(\begin{array}{c}n-1\\ k\end{array}\right)}{k+m}B(k+2m,n)\\ & =& 2\sum _{k=0}^{n-1}(-1{)}^k\frac{\left(\begin{array}{c}n-1\\ k\end{array}\right)}{k+m}\frac{(k+2m-1)!(n-1)!}{(k+2m+n-1)!}\\ & =& 2\sum _{k=0}^{n-1}(-1{)}^k\frac{\left(\begin{array}{c}n-1\\ k\end{array}\right)}{(k+m)(k+2m+n-1)\left(\begin{array}{c}k+2m+n-2\\ n-1\end{array}\right)}\\ & =& \{\frac{1}{(m+n-1)\left(\begin{array}{c}m+n-1\\ m-1\end{array}\right)}{\}}^2\end{array}$$